0
$\begingroup$

We have a call option on 3 dices with strike 3. What's the fair value of the call when it pays the min value of the 3 dices?

E.g if we throw and have 426, the min is 2 here and so call is OTM (S < K)

216 permutations (6^3)

min(6) = 1/216 x (6-3) = 0.0139 EV

(666)

min(5) = 8/215 x (5-3) = 0.0744 EV

(555 556 565 566 655 656 665 666)

min(4) = ?

I'm getting lost in the permutation formula to calculate the number of outcomes where the min. comes 4. Can someone help me figure out how to find out the correct # of outcomes with value 4?

| improve this question | |
$\endgroup$
  • $\begingroup$ 666 does not yield a minimum of 5, fyi. $\endgroup$ – Attack68♦ Feb 4 at 17:20
  • $\begingroup$ @Attack68 No but an option with a strike would be ITM if three sixes are rolled. $\endgroup$ – D Stanley Feb 4 at 18:49
2
$\begingroup$

There is a simple way to find the number of rolls with the minimum of 4:
the number of rolls with the minimum of 6 is 1.
The number of rolls with the minimum of 5 is the number of rolls for which all outcomes are 5 or 6 minus the number of rolls with the minimum of 6:
2*2*2-1=7.
The number of rolls with the minimum of 4 is the number of rolls for which all outcomes are 4,5,6 minus the number of rolls with the minimum of 5 or 6:
3*3*3-7-1=19 and so on.

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you, That's perfect! $\endgroup$ – aallove Feb 4 at 20:57
1
$\begingroup$

This isn't really an option as it is an exercise in probability.

How many rolls have a minimum of 6?

1 = 3C3 (6s)

How many rolls have a minimum of 5?

7 = 3C3 (5s) + 3C2 (5s6s) + 3C1 (5s6s)

How many rolls have a minimum of 4?

19 = 3C3 (4s) + 3C2 (4s5s) + 3C1 (4s5s) + 3C2 (4s6s) + 3C1 (4s6s) + 3P3 (4s5s6s)

How many rolls have a minimum of 3?

  3C3 (3s)
  3C2 (3s4s) 3C2 (3s5s) 3C2 (3s6s)
  3C1 (3s4s) 3C1 (3s5s) 3C1 (3s6s)
  3P3 (3s4s5s) 3P3 (3s5s6s) 3P3 (3s4s6s) = 3P3 * 3C2

= 1 + 9 + 9 + 18 = 37

How many rolls have minimum of 2?

 3C3 (2s)   = 1
 3C2 * 4C1  = 12
 3C1 * 4C1  = 12
 3P3 * 4C2  = 36

= 61

How many rolls have minimum of 1?

3C3    = 1
3C2 * 5C1 = 15
3C1 * 5C1 = 15
3P3 * 5C2 = 60

= 91

Total is 216

| improve this answer | |
$\endgroup$
  • $\begingroup$ Thanks for the breakdown! $\endgroup$ – aallove Feb 4 at 20:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.